Euclidean Geometry #1

Geometry Level pending

In the three dimensional space $R^3$ (R=set of real numbers). Find the equation of the plane passing through the point (-1,2,0) and contains the line r which is given as the intersection of these planes:

r $\equiv$ { x - 2y + z - 3 = 0, y + 3z - 5 = 0}

x + 7y + 7z -13 = 0 3x - 14y - 21z + 31 = 0 2x + y + z = 0 3x - 14y + 8z + 31 = 0 18x + 6y + 18z +18 = 0

1 solution

Guillermo Templado
Oct 21, 2015

The point doesn't belong to r. All the planes passing trough of r are y +3z - 5 = 0 and the radiation of planes (x - 2y + z - 3) + $\lambda$ (y + 3z - 5)= 0. ( $\lambda$ belonging to R) (1) As the plane passes through the point (-1,2,0), will fulfill the equation: (-1 -4 -3) + $\lambda$ (2 - 5) = 0 $\Rightarrow$ $\lambda$ = -8/3. Now, replacing in (1) we will obtain: x - 2y + z - 3 +(-8/3)(y + 3z -5)= 0 $\Rightarrow$ 3x - 6y +3z - 9 - 8y - 24z + 40 = 0 $\Rightarrow$ The equation of the plane is 3x -14y - 21z + 31 = 0 . Coming soon I'll post another solution,maybe

Euclidean Geometry #1

1 solution

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